Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202527 Apr 2025Evening ShiftMathematicsApplication of DerivativesActual

If x+ y =6, x 0, y 0 , then the maximum value of x^2 y is

Options

  1. A30
  2. B32
  3. C34
  4. D36

Correct answer

A. 30

Step-by-step solution

To maximize x^2y under the constraints x+y=6 with x 0 , y 0 , express y = 6 - x and consider the function f(x) = x^2(6-x) = 6x^2 - x^3 defined on 0 x 6 . Differentiating yields f'(x) = 12x - 3x^2 = 3x(4 - x) , which equals zero at x=0 and x=4 . Evaluating f(x) gives: f(0) = 0 , f(4) = 4^2(2) = 32 , f(6) = 0 . The maximum value is 32 . Alternatively, applying the AM-GM inequality to x 2 , x 2 , and y : x 2 + x 2 + y 3 [3] x^2y 4 Substituting x+y=6 gives 2 [3] x^2y 4 , so x^2y 32 . Equality occurs when x 2 = y and x+

Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the quadratic equation a x^2+b x+c=0 , where 2 a+3 b+6 c=0 and let g(x)= a x^3 3 + b x^2 2 +c x . Statement-I : The given quadratic equation ax ^2+ bx + c =0 has at least 2025The difference between the absolute maximum and absolute minimum values of the function f(x)=2 x^3-15 x^2+36 x-30 on [-1,4] is 2025If f(x)=x e^ x(1-x) , x R , then f(x) is 2025The angle between the curves y ^2= x and x ^2= y at the point (1,1) is 2025If the tangent of the curve 4 y^3=3 a x^2+x^3 drawn at the point (a, a) forms a triangle of area 25 24 sq.units with the coordinate axes then a = 2025If the function f(x)= x- ^2 x is defined on the interval [- , ] , then f is strictly increasing in the interval 2025If the Lagrange's mean value theorem is applied to the function f(x)=e^x defined on the interval [1,2] and the value of c (1,2) is k , then e^ k-1 = 2025If the tangent to the curve x y+a x+b y=0 at (1,1) makes an angle Tan ⁻¹ 2 with X -axis, then ab a + b = 2025 Full Application of Derivatives list All MHT CET PYQs